Question

Use the procedures developed in this chapter to find the general solution of the differential equation. 3x3y''' + 28x2y'' + 55xy' + 9y = 0

Answer 1

Substitute
`x=e^t`. By the chain rule,

`(dy)/(dx) = (dy)/(dt)\cdot(dt)/(dx)`

Now
`t=\ln(x)\implies(dt)/(dx)=\frac1x`, so

`(dy)/(dx) = \frac1x (dy)/(dt) \\\\ ~~~~~~~~ \iff (dy)/(dt) = x(dy)/(dx)`

Differentiate both sides again to recover the second and third derivatives.

`(d^2y)/(dx^2) = -\frac1{x^2}(dy)/(dt) + \frac1x (d(dy)/(dt))/(dx) \\\\ ~~~~~~~~ = -\frac1{x^2}(dy)/(dt) + \frac1x \left((d(dy)/(dt))/(dt)\cdot(dt)/(dx)\right) \\\\ ~~~~~~~~ = -\frac1{x^2} (dy)/(dt) + \frac1{x^2}(d^2y)/(dt^2) \\\\ ~~~~~~~~ \iff (d^2y)/(dt^2) - (dy)/(dt) = x^2 (d^2y)/(dx^2)`

`(d^3y)/(dx^3) = \frac2{x^3}(dy)/(dt) - \frac1{x^2}(d(dy)/(dt))/(dx) - \frac2{x^3} (d^2y)/(dt^2) + \frac1{x^2}(d(d^2y)/(dt^2))/(dx) \\\\ ~~~~~~~~ = \frac2{x^3} (dy)/(dt) - \frac1{x^2}\left((d(dy)/(dt))/(dt)\cdot(dt)/(dx)\right) - \frac2{x^3} (d^2y)/(dt^2) + \frac1{x^2}\left((d(d^2y)/(dt^2))/(dt)\cdot(dt)/(dx)\right) \\\\ ~~~~~~~~ = \frac2{x^3} (dy)/(dt) - \frac1{x^3} (d^2y)/(dt^2) - \frac2{x^3} (d^2y)/(dt^2) + \frac1{x^3}(d^3y)/(dt^3) \\\\ ~~~~~~~~ \iff (d^3y)/(dt^2) - 3 (d^2y)/(dt^2) + 2 (dy)/(dt) = x^3 (d^3y)/(dx^3)`

The ODE then transforms to a linear one,

`3\left((d^3y)/(dt^2) - 3 (d^2y)/(dt^2) + 2 (dy)/(dt)\right) + 28\left((d^2y)/(dt^2) - (dy)/(dt)\right) + 55(dy)/(dt) + 9 y = 0`

or using Lagrange's prime notation,

`3(y''' - 3y'' + 2y') + 28(y'' - y') + 55y' + 9y = 0`

`3y''' + 19y'' + 33y' + 9y = 0`

The characteristic equation is

`3r^3 + 19r^2 + 33r + 9r = (r + 3)^2 \left(r + \frac13\right) = 0`

with roots at
`r=-3` and
`r=-\frac13`, so the general solution is

`y(t) = C_1 e^(-1/3\,t) + C_2 e^(-3t) + C_3 t e^(-3t)`

Back in terms of
`x`, we get

`\boxed{y(x) = C_1 x^(-1/3) + C_2 x^(-3) + C_3 x^(-3)\ln(x)}`